1035 Password

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SoloGala

最新推荐文章于 2025-05-23 12:17:19 发布

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分类专栏： PTA 笔记文章标签： PTA A1035

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本文介绍了一个编程挑战，旨在解决由易混淆字符导致的密码问题。挑战要求将特定字符（如1,0,l,O）替换为不易混淆的符号（@,%L,o），并提供了一个C++实现示例，展示了如何检查和修改密码。

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1035 Password （20 point(s)）

To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (one) from l (L in lowercase), or 0 (zero) from O (o in uppercase). One solution is to replace 1 (one) by @, 0 (zero) by %, l by L, and O by o. Now it is your job to write a program to check the accounts generated by the judge, and to help the juge modify the confusing passwords.

题意

有一些很难区分，1 和L 0和O等。

replace 1 (one) by @, 0 (zero) by %, l by L, and O by o.

没别的。注意结果输出的时候 is 和are 还有单复数

if(CNT==1)
    cout<<"There is "<<CNT<<" account and no account is modified"<<endl;
else
    cout<<"There are "<<CNT<<" accounts and no account is modified"<<endl;

#include <iostream>
#include <string>
#include <vector>
using namespace std;
struct Node{
    string name;
    string pwd;
};
int CNT;
vector<Node> res;
bool replace(string &s){
    bool ret =false;
    for(int i=0;i<s.size();i++){
        if(s[i]=='1'){s[i]='@';ret =true;}
        else if(s[i]=='0'){s[i]='%';ret =true;}
        else if(s[i]=='l'){s[i]='L';ret =true;}
        else if(s[i]=='O'){s[i]='o';ret =true;}
    }
    return ret;
}
int main(){

    cin>>CNT;
    string temp_name,temp_pwd;
    for(int i=0;i<CNT;i++){
        cin>>temp_name>>temp_pwd;
        if(replace(temp_pwd)){
            res.push_back(Node{temp_name,temp_pwd});
        }
    }
    if(res.size()){
        cout<<res.size()<<endl;
        for(int i=0;i<res.size();i++){
            cout<<res[i].name<<" "<<res[i].pwd<<endl;
        }
    }
    else{
        if(CNT==1)
            cout<<"There is "<<CNT<<" account and no account is modified"<<endl;
        else{
            cout<<"There are "<<CNT<<" accounts and no account is modified"<<endl;
        }
    }
    return 0;
}